Problem: $\dfrac{d}{dx}\left(\dfrac{e^x}{\cos(x)}\right)=$
Solution: $\dfrac{e^x}{\cos(x)}$ is the quotient of two, more basic, expressions: $e^x$ and $\cos(x)$. Therefore, the derivative of the expression can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\dfrac{e^x}{\cos(x)}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(e^x)\cos(x)-e^x\dfrac{d}{dx}(\cos(x))}{(\cos(x))^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{e^x\cdot \cos(x)-e^x\cdot (-\sin(x))}{\cos^2(x)}&&\gray{\text{Differentiate }e^x\text{ and }\cos(x)} \\\\ &=\dfrac{e^x(\cos(x)+\sin(x))}{\cos^2(x)}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left(\dfrac{e^x}{\cos(x)}\right)=\dfrac{e^x(\cos(x)+\sin(x))}{\cos^2(x)}$ or any other equivalent form.